This page 10th CBSE math solution for exercise 7.1 part 2 is going to provide you solution for every problems that you find in the exercise no 7.1

(4) Check whether (5,-2) (6,4) and (7,-2) are the vertices of an isosceles triangle.

Solution:

Let the given points as A(5,-2) B(6,4) and C(7,-2)

**Distance between two points = √(x₂ - x₁)
² + (y₂ - y₁) ² **

Length of the side AB

Here x₁ = 5, y₁ = -2, x₂ = 6 and y₂ = 4

= **√(6 - 5)² + (4 - (-2))² **

= **√1² + (4+2)² **

= **√1 + 36**

= **√37**

Length of the side BC

Here x₁ = 6, y₁ = 4, x₂ = 7 and y₂ = -2

= **√(7 - 6)² + (-2 - 4)² **

= **√1² + (-6)² **

= **√1 + 36**

= **√37**

Length of the side CA

Here x₁ = 7, y₁ = -2, x₂ = 5 and y₂ = -2

= **√(-5 - 7)² + (-2 - (-2))² **

= **√(-12)² + (-2 + 2)² **

= **√144 + 0**

= **√144**

** = 12**

**AB = BC. Since length of two sides are equal. So it forms an isosceles triangle.**

In the page 10th CBSE math solution for exercise 7.1 part 2 we are going to see the solution of next problem

(5) In a classroom 4 friends are seated at the points A,B,C and D as shown in fig 7.8.Champa and Chameli walk into the class and after observing for a few minutes champa asks Chameli,"Don't you think ABCD is a square?" Chameli disagrees. Using distance formula, find which of them is correct.

Solution:

It can be observed that A (3,4) , B (6,7), C(9,4) and D(6,1) are the position of these 4 friends.

Length of AB

Here x₁ = 3, y₁ = 4, x₂ = 6 and y₂ = 7

= **√(6 - 3)² + (7 - 4)² **

= **√(3)² + (3)² **

= **√9 + 9**

= **√18**

** = 3****√2**

Length of BC

Here x₁ = 6, y₁ = 7, x₂ = 9 and y₂ = 4

= **√(9 - 6)² + (4 - 7)² **

= **√(3)² + (-3)² **

= **√9 + 9**

= **√18**

** = 3****√2**

Length of CD

Here x₁ = 9, y₁ = 4, x₂ = 6 and y₂ = 1

= **√(6 - 9)² + (1 - 4)² **

= **√(-3)² + (-3)² **

= **√9 + 9**

= **√18**

** = 3****√2**

Length of DA

Here x₁ = 6, y₁ = 1, x₂ = 3 and y₂ = 4

= **√(3 - 6)² + (4 - 1)² **

= **√(-3)² + (3)² **

= **√9 + 9**

= **√18**

** = 3****√2**

Length of diagonal AC

Here x₁ = 3, y₁ = 4, x₂ = 9 and y₂ = 4

= **√(9 - 3)² + (4 - 4)² **

= **√6² + 0² **

= **√36**

= **6**

Length of diagonal BD

Here x₁ = 6, y₁ = 7, x₂ = 6 and y₂ = 1

= **√(6 - 6)² + (1 - 7)² **

= **√0² + (-6)² **

= **√36**

= **6**

It can be observed that all sides of this quadrilateral ABCD are the same length and also the diagonals are of the same length.

Therefore,ABCD is a square and hence,Champa was correct.

10th CBSE math solution for exercise 7.1 part 2

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